RIP Distance vector protocol

Hi Rene,

I have a doubt in the routing tables of RIP.
In the four scenario, when the Fa1/0 link of R3 goes down, once the R2 sends a copy of its full routing table to R3, how the metric has become 2. Could you please explain it?

Regards,
Anu

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Could you post the link to the lesson you are talking about?

Hello Anusha

The scenario you’re talking about, if I’m not mistaken, is found in the following lesson:
https://networklessons.com/rip/rip-distance-vector-routing-protocol/
Specifically, it has to do with the following image:
image

So here we see an example of what would happen if the split horizon rule was not used. Just as a reminder, the split horizon rule states that if a route to a destination is learned from an advertisement that comes in from a specific interface, that route is never advertised back out of that same interface. The example shown in the mentioned lesson explains why.

So getting back to your question, R3 has lost its connection to the 3.3.3.0 network which was directly connected to its Fa0/1 interface. R3 has already told R2 that in order to reach 3.3.3.0, any packets destined for this network should be sent to itself. So, 3.3.3.0 is already in the R2 routing table.

So, R3 says it lost its connection to the 3.3.3.0 network. It sends a query to R2 and asks if it knows of an alternate route to that network. R2 indeed does have the 3.3.3.0 network in its routing table, with R3 as its next hop and a metric of 1. So R2 says to R3 “I know how to get there! Send me the packet and I’ll send it along its way” So R3 places the 3.3.3.0 network in its routing table and adds 1 to the metric advertised by R2 which is 1. So the metric becomes 1 + 1 = 2.

I hope this has been helpful!

Laz